=x2-（y+1）x+y2-y+1
=(x?
| y+1 |
| 2 |
| 1 |
| 4 |
=(x?
| y+1 |
| 2 |
| 3 |
| 4 |
∴x2+y2≥xy+x+y-1
試比較x2+y2與xy+x+y-1的大小．
試比較x2+y2與xy+x+y-1的大小．
數(shù)學(xué)人氣：834 ℃時(shí)間：2019-12-25 10:36:15
優(yōu)質(zhì)解答
∵x2+y2-（xy+x+y-1）
=x2-（y+1）x+y2-y+1
=(x?
)2?
(y+1)2+y2?y+1
=(x?
)2+
(y?1)2≥0（10分）
∴x2+y2≥xy+x+y-1
=x2-（y+1）x+y2-y+1
=(x?
=(x?
∴x2+y2≥xy+x+y-1
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